Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-2q^2 - 20q}{4q^2 + 80q + 400} \div \dfrac{q^2 - 9q}{5q^2 - 35q - 90} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2q^2 - 20q}{4q^2 + 80q + 400} \times \dfrac{5q^2 - 35q - 90}{q^2 - 9q} $ First factor out any common factors. $p = \dfrac{-2q(q + 10)}{4(q^2 + 20q + 100)} \times \dfrac{5(q^2 - 7q - 18)}{q(q - 9)} $ Then factor the quadratic expressions. $p = \dfrac {-2q(q + 10)} {4(q + 10)(q + 10)} \times \dfrac {5(q - 9)(q + 2)} {q(q - 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-2q(q + 10) \times 5(q - 9)(q + 2) } { 4(q + 10)(q + 10) \times q(q - 9)} $ $p = \dfrac {-10q(q - 9)(q + 2)(q + 10)} {4q(q + 10)(q + 10)(q - 9)} $ Notice that $(q + 10)$ and $(q - 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-10q(q - 9)(q + 2)\cancel{(q + 10)}} {4q\cancel{(q + 10)}(q + 10)(q - 9)} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $p = \dfrac {-10q\cancel{(q - 9)}(q + 2)\cancel{(q + 10)}} {4q\cancel{(q + 10)}(q + 10)\cancel{(q - 9)}} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $p = \dfrac {-10q(q + 2)} {4q(q + 10)} $ $ p = \dfrac{-5(q + 2)}{2(q + 10)}; q \neq -10; q \neq 9 $